NCERT Exemplar Class 11 Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations

# NCERT Exemplar Class 11 Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations

Edited By Ravindra Pindel | Updated on Sep 12, 2022 05:49 PM IST

NCERT Exemplar Class 11 Maths solutions chapter 5 covers Complex numbers and its applications. Complex numbers are something which we have been studying over the years with any ‘real number’ being said to be a complex number, which can be denoted as a variable or alphabet in simpler terms. The other number, along with the real number, is a part of the complex number, is called the ‘imaginary number.’ Both of them appear together in a problem. Terms such as integer, conjugate, square root, polar, etc. will be studied in NCERT Exemplar Class 11 Maths solutions chapter 5, that will be utilised to find results for these complex number problems.

## NCERT Exemplar Class 11 Maths Solutions Chapter 5: Exercise - 1.3

Question:1

Answer:

$=(1-i)^n (1+i )^n$

Question:3

If then find (x, y).

Answer:

Given $\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3= x+iy$
$=\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3$
$=\left (\frac{1+2i+i^2}{1-i^2} \right )^{3}-\left (\frac{1-2i+i^2}{1-i^2} \right )^{3}$

Question:5

If $\left(\frac{1-i}{1+i}\right)^{100}=a+ib$ then find (a, b).

Answer:

$a+ib=\left(\frac{1-i}{1+i}\right)^{100}=\left [\frac{1-i}{1+i}*\frac{1-i}{1-i} \right ]^{100}$

Question:6

If , find the value of .

Answer:

Question:7

If then show that $z= -i\bar{z}$

Answer:

Rationalising the denominator,

Hence,Proved.

Question:8

If, then show that where b?R, representing z in the complex plane is a circle.

Answer:

This equation is a equation of circle

Question:9

If the real part of is 4, then show that the locus of the point representing z in the complex plane is a circle.

Answer:

$real \: \: part=4 \Rightarrow \frac{(x-1)(x+2)+y^2}{(x-1)^2+y^2}=4$

The equation obtained is that of a circle. Hence, locus of z is a circle.

Question:10

Show that the complex number z, satisfying the condition lies on a circle.

Answer:

Let z=x+iy

$\tan^{-1}\frac{y}{x-1}-\tan^{-1}\frac{y}{x+1}=$

$\tan^{-1}\left(\frac{y\left ( x+1+-x+1 \right )}{x^2-1+y^2}\right) = \frac{\pi}{4}$
$\tan\frac{\pi}{4} = \frac{2y}{x^2+y^2-1}$

The equation obtained represents the equation of a circle

Question:11

Solve that equation .

Answer:

Comparing real and imaginary parts
And
Substituting the value of y in

Hence,
Hence,

Question:12

If , then find z

Answer:

We have
Substituting we get
$|z|= \sqrt{(x^2+y^2 ) }=\sqrt{(x+1)^2+y^2 } = (x+2)+i(y+2)$
Comparing real and imaginary parts, $\sqrt{(x+1)^2+y^2 } = (x+2)$

Substituting the value of y in $\sqrt{(x+1)^2+y^2 } = (x+2)$

Hence,
Thus, $z=x+iy=\frac{1}{2}-2i$

Question:13

If , then find x – 1 : y. where

Answer:

Given that

$arg(x-1+iy)=arg(x+i(y+3))$

Question:14

Show that represents a circle. Find its centre and radius.

Answer:

Substituting , we get

$x^2+y^2-\frac{20}{3} x+\frac{32}{}3=0$
$(x-\frac{10}{3})^2+y^2+\frac{32}{3}-\frac{100}{9}=0$
Thus, the centre of circle is $(\frac{10}{3},0)$ and radius is

Question:15

If is a purely imaginary number, then find the value of .

Answer:

Let

is purely imaginary

Hence,

Question:16

z1 and z2 are two complex numbers such that and , then show that .

Answer:

Let $z_1=\left |z_1 \right |\left (cos\theta _1+i\ sin\theta_1 \right )$ and $z_2=\left |z_2 \right |\left (cos\theta _2+i\ sin\theta_2 \right )$
Given that
And

$z_2=\left |z_2 \right |\left (cos\theta _2+i\ sin\theta_2 \right )$

$z_1=|z_2 |\left ( -\cos \theta_2+isin\theta_2 \right )$

$z_1=-|z_2 |\left ( \cos \theta_2-isin\theta_2 \right )$

$z_1=-|z_2 |\left ( \cos \theta_2-isin\theta_2 \right )$,

Question:17

If |z1| = 1 (z1 ≠ –1) and then show that the real part of z2 is zero

Answer:

Let

Since,

Therefore, the real part of is zero

Question:18

If z1, z2 and z3, z4 are two pairs of conjugate complex numbers, then find

Answer:

and are conjugate complex numbers.
The negative side of the real axis

Similarly,

Question:19

If , then
Show that $|z_1 + z_2 + z_3 +... + z_n| =|1/z_1 +1/z_2 +...+1/z_n |$

Answer:

Now,
=$\left | \frac{1}{\bar{z_1}}+\frac{1}{\bar{z_2}}+...+\frac{1}{\bar{z_n}} \right |$

Question:21

Solve the system of equations $Re (z^2) = 0, |z| = 2.$

Answer:

$Re (z^2) = 0, |z| = 2.$

Hence,

Question:22

Find the complex number satisfying the equation .

Answer:

Substituting we get

1+y = 0
y = -1

Hence,

Question:23

Write the complex number in polar form

Answer:

$=\sqrt2\left [ \cos \left ( -\frac{\pi}{4}-\frac{\pi}{3}\right )+i\sin\left (-\frac{\pi}{4}-\frac{\pi}{3} \right ) \right ]$

$=-\sqrt2\left [ \cos \left ( \frac{5\pi}{12}\right )+i\sin\left (\frac{5\pi}{12} \right ) \right ]$

Question:24

If z and w are two complex numbers such that and , then show that .

Answer:

Let $z=\left | z \right |(\cos\theta_1 + i sin\theta_1 )$ and $w=\left | w \right |(\cos\theta_2 + i sin\theta_2 )$

Also

$\bar{z} w=|z|\left (\cos \theta _{1}-i\sin\theta _{1} \right )w$
$w=|w|\left (\cos \theta _{2}+i\sin\theta _{2} \right )=1$
$\bar{z}w$
$\bar{z}w$

Question:25

i) For any two complex numbers z1, z2 and any real numbers a, b,

ii) The value of $\sqrt{-25} * \sqrt{-9}$ is

iii) The number $\frac{(1-i)^3}{1-i^3}$ is equal to .......

iv) The sum of the series $i+i^2+i^3+...+i^{1000}$ upto 1000 terms is ...

v) Multiplicative inverse of 1 + i is ................

vi) If $z_1 \ and \ z_2$ are complex numbers such that $z_1 + z_2$ is a real number, then $z_2$ =

vii) $(\bar{z}\neq0)$ is ....

viii) If $|z+4|\leq 3$then the greatest and least values of |z+1| are ..... and .....

ix) if then the locus of z is .......

x) If $|z|=4$ and then z=

Answer:

i)

ii)
iii)

iv)
v) $\frac{1}{1+i}=1*\frac{1-i}{(1+i)(1-i)} =\frac{1}{2} (1-i)$
vi) Let

If is real then

So,
vii)

viii) $|z+4|\leq 3$

The gratest value is 6 and the least value of is 0
ix)
Let

which represents an equation of a circle
x) $|z|=4$ and
Let

Question:26

State True of False for the following:

i) The order relation is defined on the set of complex numbers.

ii) Multiplication of a non-zero complex number by -i rotates the point about origin through a right angle in the anti-clockwise direction
iii) For any complex number z, the minimum value of |z| + |z – 11 is 1
iv) The locus represented by |z — 11= |z — i| is a line perpendicular to the join of the points (1,0) and (0, 1)
v) If z is a complex number such that z ≠ 0 and Re(z) = 0, then Im (z2) = 0
vi) The inequality |z – 4| < |z – 2| represents the region given by x > 3.
(vii) Let Z1 and Z2 be two complex numbers such that |z, + z2| = |z1 j + |z2|, then arg (z1 – z2) = 0.
(viii) 2 is not a complex number.

Answer:

(i) Comparison of two purely imaginary complex numbers is not possible. However, the two purely real complex number can be compared. So, it is false.

(ii) Let

which rotates at angle of 180. So, it is ‘false’.

(iii) Let

The value of is minimum when

Hence, it is true.

iv) Let

Given that

which is a straight line slope=1

Now, equation of line through the point 1,0and 0,1

whose slope=-1

Multiplication of the slopes of two lines =-1*1=-1

So, they are perpendicular. Hence, it is true.

v)Let

Since, real part is 0

which is real Hence, it is False.

vi)

Let

Hence, it is true.

vii) Let and

$|z_1+z_2 |=|z_1 |+|z_2 |$

$|x_1+iy_1+x_2+iy_2 |=|(x_1+iy_1 )+(x_2+y_2 i)|$

Squaring both sides, we get )

Again squaring on both sides we get

Hence, it is false.

(viii) Since, every real number is a complex number. So, 2 is a complex number. Hence, it is false.

Question:27

 COLUMN A COLUMN B a) The polar form of is i) Perpendicular bisector of segment joining (-2,0) and (2,0) b)The amplitude of is ii) On or outside the circle having centre at (0,-4) and radius 3. c) If then locus of z is iii) d)If then locus of z is iv)Perpendicular bisector of segment joining (0,-2) and (0,2) e) Region represented by is v) f) Region represented by is vi) On or outside the circle having centre at (-4,0) and radius 3units. g)Conjugate of lies in vii) First Quadrant h) Reciprocal of 1-i lies in viii) Third Quadrant

Answer:

a)Given

Polar form of z .

And

Since

Polar form of z

b) Given that

Here argument z

So,

Since, x<0 and y>0

c)

which represents equation of-y axis and is perpendicular to the line joining the points (-2,0) and (2,0)

d)

which is the equation of x-axis and is perpendicular to the line segment joining (0,-2)and (0,2)

e)

which represents a circle on or outside having centre (0,-4)

f)

Let Then,

which is a circle having centre (-4,0) and and is on the circle.

g) Let

which lies in third quadrant..

h) Given that z=1-i

Reciprocal of

which lies in first quadrant.

Question:28

What is the conjugate of ?

Answer:

Hence,

Question:29

If , is it necessary that ?

Answer:

Let

$\\Consider z_1 = 3 + 4i \\ z_2 = 4 + 3i$
Hence, it is not necessary that

Question:31

Find z if and ..

Answer:

Polar form of

Question:32

Find .

Answer:

$\left | (1+i)\frac{(2+i)}{(3+i)}*\frac{3-i}{3-i} \right |=\left | (1+i).\frac{6-2i+3i-i^2}{9-i^2} \right |$
$=\left | \frac{(1+i)(7+i)}{9+1} \right |$

Question:33

Find principal argument of .

Answer:

Now Re(z)<0 and image(z)>0 $arg(z)=\pi-\alpha=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$

Question:34

Where does z lie, if $|\frac{z-5i}{z+5i}|=1$.

Answer:

Hence, z lies on x-axis i.e., real axis.

Question:35

$\sin x + i \cos 2x$ and $\cos x - i \sin 2x$ are conjugate to each other for:
A. $x = n\pi$

B.

C.
D. No value of x

Answer:

The answer is the option (c).

Comparing the real and imaginary part, we get

$\tan \mathrm{x}=1 \text { and } \tan 2 \mathrm{x}=1$
$\tan 2 \mathrm{x}= \frac{2\tan x }{1-\tan^2 x}=1$
The above value is not satisfied by tan x = 1. Hence no value of x is possible.

Question:36

The real value of α for which the expression is purely real is:
A.
B.
C.
D. None of these

Answer:

The answer is the option (c).
Let
\begin{aligned} &\begin{array}{c} =\frac{(1-i\sin \alpha)(1-2 i \sin \alpha)}{(1+2 i \sin \alpha)(1-2 i \sin \alpha)} \\\\ =\frac{1-2 i \sin \alpha-i \sin \alpha+2 i^{2} \sin ^{2} \alpha}{(1)^{2}-(2 i \sin \alpha)^{2}} \\\\ =\frac{1-3 i \sin \alpha-2 \sin ^{2} \alpha}{1-4 i^{2} \sin ^{2} \alpha}=\frac{\left(1-2 \sin ^{2} \alpha\right)-3 i \sin \alpha}{1+4 \sin ^{2} \alpha} \\\\ \frac{-3 i \sin \alpha}{1+4 \sin ^{2} \alpha}=0 \\\\ \sin \alpha=0 \quad \operatorname{} \alpha=n \pi \end{array}\\ &\text { Hence, c is correct. } \end{aligned}

Question:37

If z = x + iy lies in the third quadrant, the also lies in the third quadrant if
A. x > y > 0
B. x < y < 0
C. y < x < 0
D. y > x > 0

Answer:

The answer is the option (b).
If z lies in the third quadrant,So,$x<0 \: \: and\: \: y<0\: \: , z=x+iy$

When z lies in third quadrant then will also be in third quadrant.
$\begin{array}{c} \frac{x^{2}-y^{2}}{x^{2}+y^{2}}<0 \text { and } \frac{2 x y}{x^{2}+y^{2}}>0 \\\\ x^{2}-y^{2}<0 \text { and } 2 x y>0 \\\\ x^{2}0 \\\\ \text { So, } x
Hence, b is correct..

Question:38

The value of is equivalent to
A.
B. |z – 3|
C.
D. None of these

Answer:

The answer is the option (a).

Question:39

If , then
A.
B.
C.
D.

Answer:

The answer is the option(b).
$\begin{array}{c} \left(\frac{1+i}{1-i}\right)^{x}=1 \\\\ \left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)^{x}=1 \\\\ \left(\frac{1-1+2 i}{1+1}\right)^{x}=1 \\\\ \left(\frac{2 i}{2}\right)^{x}=1(i)^{x}=1 \end{array}$
$x = 4 n, n \epsilon N$
Hence, the correct option is

Answer:

The answer is the option (a).

Multiplying eqn (i )and (ii )we get

Question:41

Answer:

The answer is the option (a).
$\begin{array}{c} z_{1}=r_{1}\left(\cos \theta_{1}+i \sin \theta_{2}\right) \\ \left|z_{1}\right|=r_{1} \\ z_{2}=r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right) \\ \left|z_{2}\right|=r_{2} \\\\ z_1z_2=r_{1} r_{2}\left[\cos \left(\theta_{1}+\theta_{2}\right)+i \sin \left(\theta_{1}+\theta_{2}\right)\right] \\ =r_{1}\left(\cos \theta_{1}+i \sin \theta_{1}\right) \cdot r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right) \\ \end{array}$

Hence, option a is correct.

Question:42

Answer:

The answer is the option (b).
If z rotated through an angle of about the origin in clockwise direction .
Then the new position $=z.e^{-\left ( \frac{\pi}{2} \right )}$
$=\left (2-i \right ).e^{-\left ( \frac{\pi}{2} \right )}$
$=\left (2-i \right ).\left [ \cos\left (- \frac{\pi}{2} \right )+isin\left (- \frac{\pi}{2} \right ) \right ]$
$=\left (2-i \right ).\left (0-i \right )$

Hence, the correct option is (b)

Question:43

Answer:

The answer is the option (d).
is a non real complex number if ,

Hence, d is correct

Question:44

If , then
A.
B.
C.
D.

Answer:

The answer is the option (d).

Squaring both sides,
Hence, d is correct

Question:45

The complex number z which satisfies the condition lies on
A. circle
B. the x-axis
C. the y-axis
D. the line

Answer:

The answer is the option (b).

Hence, b is correct

Question:46

If z is a complex number, then
A.
B.
C.
D.

Answer:

. The answer is the option (b).

Hence, b is correct

Question:47

B. z2=z1
C.
D.

Answer:

The answer is the option (c).
Since,
$|z_1+z_2 |=|r_1 \left ( \cos \theta _1+i\sin\theta _1 \right )+r_2 \left ( \cos \theta _2+i\sin\theta _2 \right )|$

So,
Squaring both sides,

Question:48

B.
C.
D. none of these

Answer:

On solving we get

If z is real number then

$\\3 \cos \theta=0 \\\cos \theta=0$

Question:49

The value of when is:
A. 0
B.

C. π
D. none of these

Answer:

The answer is the option (c).
$\\ Let\: \: z= -x+0i \: \: and \: \: x<0 \\ |z|=\sqrt{(-1)^2+(0)^2 }=1,x<0$

Since, the point (-x,0)lies on the negative side of the real axis , principal argument (z)=π

Hence, option c is correct

Question:50

B.
C.
D. none of these

Answer:

The answer is the option (a).
Given that $\\ z=1+2i \\ |z|=\sqrt{(1)^2+(2)^2 }=\sqrt5$
Now

\begin{aligned} =\frac{3-i}{2-2 i} * \frac{2+2 i}{2+2 i}=\frac{6+6 i-2 i-2 i^{2}}{4-4 i} \\ =\frac{6+4 i+2}{4+4} &=\frac{8+4 i}{8} \\ =1+\frac{1}{2} i & \end{aligned}

Hence, option a is correct

## Important Notes from NCERT Exemplar Class 11 Maths Solution chapter 5

Various laws need to be followed and derived by a mathematician when it comes to multiplying the two complex numbers. All of them are named over the features of the specific formula derivation. Ahead, you will also learn about the Modulus and Conjugate of complex numbers and how to solve sums with their help. There is an Argand and Polar representation in the chapter, relating to graphical presentations. A point needs to be derived at the ‘x’ and ‘y’ axis to form an argand or complex plane.

Students, with the help of NCERT Exemplar Class 11 Maths solutions chapter 5, will not face any issues when trying to solve such problems. It will also help them score better in exams.

All the concepts have been covered in NCERT Exemplar solutions for Class 11 Maths chapter 5. By using NCERT Exemplar Class 11 Maths chapter 5 solutions PDF Download function, students can access quality study material that is effectively constructed by experts for the best learning experience.

## Main Subtopics in NCERT Exemplar Class 11 Maths Solution Chapter 5

• Introduction
• Complex Numbers
• Algebra of Complex Numbers
• Addition of two complex numbers
• Difference of two complex numbers
• Multiplication of two complex numbers
• Division of two complex number
• Power of i
• The square roots of a negative real number
• Identities
• The Modulus and the Conjugate of a Complex Number
• Argand Plane and Polar Representation
• polar representation of a complex number
• Quadratic Equations

## What will the students learn from NCERT Exemplar Class 11 Maths Solution Chapter 5?

Belonging to the algebraic part of the subject, it is always fun for the students to solve these problems, once they are clear with the concept. NCERT Exemplar Class 11 Maths chapter 5 solutions will help in understanding the said concept and give examples in how to apply them to questions of any kind.

Other than that, in geometric applications, complex numbers are also very useful, for students in their near future. Students, through NCERT Exemplar Class 11 Maths solutions chapter 5, will also learn graphs in an advanced manner with newer placements and some difficult problems, which would make them sharper and clever.

## NCERT Solutions for Class 11 Mathematics Chapters

 Chapter 1 Sets Chapter 2 Relations and Functions Chapter 3 Trigonometric Functions Chapter 4 Principles of Mathematical Induction Chapter 6 Linear Inequalities Chapter 7 Permutations and Combinations Chapter 8 Binomial Theorem Chapter 9 Sequences and Series Chapter 10 Straight lines Chapter 11 Conic Sections Chapter 12 Introduction to Three Dimensional Geometry Chapter 13 Limits and Derivatives Chapter 14 Mathematical Reasoning Chapter 15 Statistics Chapter 16 Probability

## Important Topics in NCERT Exemplar Class 11 Maths Solutions Chapter 5

· Class 11 Maths NCERT Exemplar solutions chapter 5 has detailed that complex numbers, their operations, square root, power, identities, representation of complex numbers and quadratic equations are important topics which students should pay extra attention to.

· In NCERT Exemplar Class 11 Maths chapter 5 solutions, students will also learn about quadratic equations. A quadratic equation is an equation where rearrangement of a particular equation is done via a standard form, where x is the unknown number, whereas, a, b and c, are known numbers. An equation with four numbers is said to be quadratic equations.

Check Chapter-Wise NCERT Solutions of Book

 Chapter-1 Sets Chapter-2 Relations and Functions Chapter-3 Trigonometric Functions Chapter-4 Principle of Mathematical Induction Chapter-5 Complex Numbers and Quadratic equations Chapter-6 Linear Inequalities Chapter-7 Permutation and Combinations Chapter-8 Binomial Theorem Chapter-9 Sequences and Series Chapter-10 Straight Lines Chapter-11 Conic Section Chapter-12 Introduction to Three Dimensional Geometry Chapter-13 Limits and Derivatives Chapter-14 Mathematical Reasoning Chapter-15 Statistics Chapter-16 Probability

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3. Which topics are added in this chapter?

In this chapter, the complex numbers are explained, their properties, operations on complex numbers along with quadratic equations.

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